A conducting spherical shell having an inner radius of 3 cm and outer radius of 4.3 cm carries a net charge of 9.6 μC. A conducting sphere is placed at the center of this shell hav- ing a radius of 0.8 cm and carries a net charge of 1.7 μC. Determine the surface charge density on the outer surface of the shell.

sigma = 3.4*10^-4 C / m^2

Explanation:

Given:

- The net charge of the conducting sphere Q_s = 1.7 uC

- The net charge in the shell Q_shell = 9.6 uC

- Outer radius of the shell r = 4.3 cm

Find:

Determine the surface charge density on the outer surface of the shell.

Solution:

- The concept here is that the net charge on the conducting sphere must be equal to charge on inner shell. Hence, Q_s = Q_in

                                   Q_shell = Q_in + Q_out

                                   Q_out = Q_shell - Q_in

                                   Q_out = 9.6 - 1.7 = 7.9 uC

- Now we can calculate the charge density of the outer shell by:

                                   sigma = Q_out / A_out

                                   sigma = (7.9*10^-6) / 4*pi*(0.043)^2

                                  sigma = 3.4*10^-4 C / m^2

answer:

yes

explanation:

answer: report them

explanation: